TITLE: When does $(e^a)^b = e^{ab}$ hold?
QUESTION [9 upvotes]: For a complex number $A$ and a real number $B$, when does the well-known formula $(e^A)^B = e^{AB}$ fail? Or does it hold at all for complex A?   
Since $e^{2\pi i} = 1$, if this formula holds for all complex numbers $A$ and real numbers $B$, then it would imply that $e^{2\pi ti}=( e^{2\pi i} )^t = 1^t=1$ for all $t$, which is obviously false.

REPLY [3 votes]: By definition, 
$$ \exp(A)^B = \exp(B \ln(\exp(A))) = \exp(B (A + 2 \pi i n)) = \exp(AB) \exp(2\pi i n B)$$
where $n$ is an integer.  Different values of $n$ determine which branch of the logarithm (and thus of the power) is being used.  In particular, $\exp(2 \pi i n B) = 1$ if and only if $nB$ is an integer. 
In particular, suppose you're using the principal branch, i.e. $-\pi < \text{Im}(\log(z)) \le \pi$ for all $z$.  If $A = x + i y$, $\ln(\exp(A)) = x + i y + 2 \pi i n$ where $n$ is chosen so that $-\pi < y + 2 \pi n \le \pi$.  In particular
$n = 0$ if and only if $-\pi < \text{Im} A \le \pi$.  For such $A$, your identity is true for all complex $B$.  If $\text{Im} A$ is not in that interval, $n \ne 0$ and your identity is only true for a discrete set of values of $B$.